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KULIAH V & VII
MEKANIKA TEKNIK TI
BEAMGAYA INTERNAL, DIAGRAM
GAYA GESER DAN MOMEN
OLEH:
ALIEF WIKARTA, ST
JURUSAN TEKNIK MESIN
FTI ITS SURABAYA, 2007
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Defii!i Be"#
Beam- structural member designed to support
loads applied at various points along its length.
Beam designis two-step process:
1) determine shearing forces and bending
moments produced by applied loads2) select cross-section best suited to resist
shearing forces and bending moments
Beam can be subjected to concentratedloads or
distributedloads or combination of both.
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A$" i% '"(" I%e)"* +
traight two-force memberABis in
e!uilibrium under application of Fand
-F. Internal forcese!uivalent toFand -Fare
re!uired for e!uilibrium of free-bodies
ACand CB.
'"(" I%e)"* : "(" (" #ei-"% .e)!"#" .e).""i."i" !%)-%) !e/i" !%)-%) %e)!e.% #e"1i --/
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Re"-!i $"1" Be"#
Beams are classified according to way in which theyare supported.
"eactions at beam supports are determinate if they
involve only three un#nowns. $therwise% they are
statically indeterminate.
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'"(" 'e!e) 1" M#e $"1" Be"#
&ish to determine bending moment
and shearing force at any point in a
beam subjected to concentrated and
distributed loads.
'etermine reactions at supports by
treating whole beam as free-body.
(ut beam at Cand draw free-body
diagrams forACand CB. By
definition% positive sense for internal
force-couple systems are as shown.
rom e!uilibrium considerations%
determineM and VorMand V.
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Di")"# '"(" 'e!e) 1" M#e $"1" Be"#
*ariation of shear and bending
moment along beam may beplotted.
'etermine reactions at
supports.
(ut beam at Cand consider
memberAC%22 PxMPV +=+=
(ut beam atEand consider
memberEB%
( ) 22 xLPMPV +== or a beam subjected to
concentrated loads% shear is
constant between loading points
and moment varies linearly.
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3%/ S"* 4
'raw the shear and bending moment
diagrams for the beam and loading
shown.
$+,$/:
a#ing entire beam as a free-body%
calculate reactions atBandD.
ind e!uivalent internal force-couplesystems for free-bodies formed by
cutting beam on either side of load
application points.
0lot results.
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3%/ S"* 4 5"6"."
0lot results.
/ote that shear is of constant value
between concentrated loads and
bending moment varies linearly.
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3%/ S"* 2
Given: beam is supported
by a hinge at % a roller at B.
orce applied at (. oment
applied at '.
Find: 'raw the shear and
bending moment diagrams
Plan:
a) 'raw a B' of the beam.
b)(alculate support reactions.
c) ind e!uivalent internal force-couple systems for free-bodies formed
by cutting beam on either side of load application points.
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3%/ S"* 2 5"6"."
3y4 5
y6 By7 855 4 5
5.1 6 By7 5.8 4 5
By4 5.9 #ip
3B4 5
- y25) 6 5.815) 7 ; 4 5
25y4 8 - ;
y4 5.1 #ip
y By
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3%/ S"* 2 5"6"."
0otongan 1-1
31-14 5
5.1 7 * 4 5
* 4 5.1 #ip
< 4 5 ft *4 5.1 #ip
< 4 15 ft *(4 5.1 #ip
31-14 5
7 5.1
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3%/ S"* 2 5"6"."
0otongan 2-2
32-24 5
5.1 7 5.8 7 * 4 5
* 4 5.1 7 5.8 #ip
* 4 - 5.9 #ip
< 4 15 ft *(4 - 5.9 #ip
< 4 19 ft *'
4 - 5.9 #ip
32-24 5
7 5.1
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3%/ S"* 2 5"6"."
0otongan =-=
5.1 #ip 5.9 #ip 0.1 kip15 x < 20 ft
V
M
x
3=-=4 5
5.1 7 5.8 7 * 4 5
* 4 5.1 7 5.8 #ip
* 4 - 5.9 #ip
< 4 19 ft *'4 - 5.9 #ip
< 4 25 ft *'
4 - 5.9 #ip
3=-=4 5
7 5.1
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3%/ S"* 2 5"6"."
'iagram >eser dan omen
31-14 5 4 5.1
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3%/ S"* 8
'raw the shear and bending moment
diagrams for the beamAB. he
distributed load of ?255 /@m. e
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3%/ S"* 8 5"6"."
$+,$/:
a#ing entire beam as a free-body% calculatereactions atAandB.
:5= AM
( ) ( ) ( ) ( ) ( ) 5m99.5/1A55m19.5/2185mA.5 =B
/18;2=B
:5= BM
( ) ( ) ( ) ( ) ( ) 5mA.5m29.5/1A55m89.5/2185 =+ A
/2=1A=
A
:5= xF 5=xB
/ote: he 1A55 / load atEmay be replaced by
a 1A55 / force and 1A5 /m. couple atD.
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3%/ S"* 8 5"6"."
:51 =M ( ) 5?2552=1A 21
=++ Mxxx
2=8552=1A xxM =
:52 =M ( ) 519.521852=1A =++ Mxx
( )m/19A=2;
+=xM
rom CtoD:
= :5F 521852=1A = V
/19A=V
valuate e!uivalent internal force-couple systems
at sections cut within segmentsAC% CD% andDB.
rom to C:
= :5F 5?2552=1A = Vx
xV ?2552=1A=
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3%/ S"* 8 5"6"."
:52 =M
( ) ( ) 5;9.51A551A519.521852=1A =+++ Mxxx
( ) m/18;21=1; = xM
valuate e!uivalent internal force-couple
systems at sections cut within segmentsAC%
CD% andDB.
romDtoB:
= :5F 51A5521852=1A = V
/18;2=V
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3%/ S"* 8 5"6"."
0lot results.rom to C:
xV ?255=21A=
225=21A xxM =
rom CtoD:
/19A=V
( ) m/19A=2; += xM
romDtoB:/18;2=V
( ) m/18;21=1; = xM
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3%/ S"* 9
'raw the shear and bending-moment diagrams
for the beam and loading shown.
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3%/ S"* 9 5"6"."
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